Question: Define $f(x, y, z) = x^2\ln(z)$. Let $\vec{a} = (-1, 0, e)$ and $\vec{v} = \left( 1, 0, 0 \right)$. Calculate $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h}$.
Answer: When a directional derivative has a direction that equals $(1, 0, 0)$, $(0, 1, 0)$, or $(0, 0, 1)$, it becomes a regular partial derivative. Because $v = (1, 0, 0)$, the limit we want to find is the definition of $\dfrac{\partial f}{\partial x}$ evaluated at $(-1, 0, e)$. Therefore: $ \lim_{h \to 0} \dfrac{f \left( -1 + h, 0, e \right) - f(-1, 0, e)}{h} = \dfrac{\partial f}{\partial x}(-1, 0, e)$ $\begin{aligned} &\dfrac{\partial f}{\partial x} = 2x\ln(z) \\ \\ &\dfrac{\partial f}{\partial x}(-1, 0, e) = -2 \end{aligned}$ In conclusion, $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h} = -2$.